Network Fundamentals

398 questions

All types398 Multiple choice205 Multiple select92 Numeric18 Matching68 Ordering7 Short answer8

A network technician is troubleshooting interface and cabling symptoms across enterprise switches. Match each Cisco IOS interface status output or error counter on the left with its most probable root cause on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match 'down/down' to Layer 1 physical failure; 'up/down' to Layer 2 data link issue; 'Late Collisions' to duplex mismatch; and 'CRC errors' to physical media impairment or electromagnetic interference.

Each status combination and error counter directly corresponds to its underlying network layer or physical cabling cause. An interface status of 'down/down' reflects Layer 1 physical disconnection. An interface status of 'up/down' indicates Layer 2 data link failure. Late collisions indicate a duplex configuration mismatch. CRC errors without collisions indicate physical frame corruption due to cable degradation or electrical interference.

Step-by-Step Solution

Analyze interface operational states.

Identify that 'down/down' indicates a Physical Layer (Layer 1) connectivity issue, while 'up/down' signifies operational Layer 1 but failed Data Link Layer (Layer 2) keepalives or framing.

Cisco IOS interface output separates physical signaling (interface state) from data link communication (line protocol state).

Differentiate between specific interface error counters.

Link Late Collisions to duplex mismatches where collision detection window boundaries are violated, and associate CRC errors with frame corruption due to faulty cabling or interference.

Accurate error counter diagnosis isolates misconfigurations from physical cabling damage.

Key Concept

Cisco IOS Interface Status and Troubleshooting Error Counters

Question 262Question →

Which two modifications are performed on a 48-bit MAC address when generating a 64-bit IPv6 interface identifier using the EUI-64 process? (Select TWO.)

Select all that apply

The 16-bit hexadecimal value FFFE is inserted into the middle of the MAC address.

The 7th bit (Universal/Local bit) of the first byte of the MAC address is inverted.

The 16-bit hexadecimal value FFFF is inserted into the middle of the MAC address.

The first 24 bits of the MAC address are overwritten by the prefix FE80.

Show answer & explanation

Answer: The 16-bit hexadecimal value FFFE is inserted into the middle of the MAC address.; The 7th bit (Universal/Local bit) of the first byte of the MAC address is inverted.

Answer

Generating an IPv6 EUI-64 interface identifier requires inserting the 16-bit hexadecimal value FFFE into the middle of the 48-bit MAC address and inverting the 7th bit (Universal/Local bit) of the first byte.

The standard EUI-64 process creates a 64-bit interface ID from a 48-bit MAC address by inserting the 16-bit hexadecimal value FFFE between the OUI and vendor-assigned portion, and inverting the 7th bit (Universal/Local bit) of the first byte.

Step-by-Step Solution

Split the 48-bit MAC address into two 24-bit halves.

The Organizationally Unique Identifier (OUI) forms the upper 24 bits and the NIC extension forms the lower 24 bits.

EUI-64 requires extending the address length at its midpoint.

Insert FFFE between the two 24-bit halves.

The address expands from 48 bits to 64 bits.

FFFE acts as the designated padding delimiter specified by IEEE and IPv6 standards.

Invert the 7th bit (Universal/Local bit) of the first byte.

If the 7th bit was 0 (indicating a burned-in MAC address), it becomes 1 in the IPv6 interface identifier.

IPv6 standardizes global uniqueness by setting the U/L bit to 1 for universally unique interface IDs.

Key Concept

EUI-64 Interface Identifier Derivation

Question 263Question →

A network technician is connecting legacy network devices in a lab environment where Auto-MDIX is explicitly disabled on all switch interfaces. Which two statements correctly describe the cable pinout requirements and cabling types for these connections? (Choose two.)

Select all that apply

Connecting two network switches directly via FastEthernet interfaces requires a crossover cable when Auto-MDIX is disabled.

An Ethernet crossover cable connects transmit pins 1 and 2 on one RJ-45 connector to receive pins 3 and 6 on the opposite connector.

Connecting a switch port directly to a router Ethernet interface requires a crossover cable when Auto-MDIX is disabled.

A standard straight-through Ethernet cable maps pins 1 and 2 on one end to pins 3 and 6 on the opposite end.

A rollover cable must be used to connect two Ethernet switches together for standard data transmission.

Show answer & explanation

Answer: Connecting two network switches directly via FastEthernet interfaces requires a crossover cable when Auto-MDIX is disabled.; An Ethernet crossover cable connects transmit pins 1 and 2 on one RJ-45 connector to receive pins 3 and 6 on the opposite connector.

Answer

Connecting two network switches directly via FastEthernet interfaces requires a crossover cable when Auto-MDIX is disabled, and an Ethernet crossover cable connects transmit pins 1 and 2 on one RJ-45 connector to receive pins 3 and 6 on the opposite connector.

When connecting identical network device types (such as switch-to-switch or router-to-router) without Auto-MDIX feature active, a crossover cable must be used so that the transmit pins (1 and 2) on one end link to the receive pins (3 and 6) on the opposite end.

Step-by-Step Solution

Identify the device types being connected

Connecting a switch to a switch involves two like devices (MDI-X to MDI-X).

Like devices transmit and receive on identical pin pairs.

Determine the required cable type without Auto-MDIX

A crossover cable is required for switch-to-switch connections.

Without Auto-MDIX to automatically swap transmit and receive signals, the physical cable must perform the crossover function.

Verify the pinout wiring for FastEthernet crossover cables

Pins 1 and 2 on one connector connect to pins 3 and 6 on the opposite connector.

FastEthernet (100BASE-TX) uses pairs 1/2 and 3/6 for transmit (Tx) and receive (Rx).

Key Concept

Ethernet Cable Pinouts and MDI/MDI-X Device Interfaces

Question 264Question →

A network administrator is assigning IP addresses to devices within the $172.16.10.16/28$ subnet. Which TWO of the following IPv4 addresses are valid host addresses that can be assigned to network interfaces on this subnet? (Select TWO.)

Select all that apply

172.16.10.17

172.16.10.28

172.16.10.16

172.16.10.31

Show answer & explanation

Answer: 172.16.10.17; 172.16.10.28

Answer

The IPv4 addresses 172.16.10.17 and 172.16.10.28 are valid assignable host addresses.

For the prefix

172.16.10.16/28

, the subnet mask is

255.255.255.240

, giving a subnet block size of

16

. The subnet range spans from

172.16.10.16

172.16.10.31

. Within this subnet,

172.16.10.16

is the reserved network ID and

172.16.10.31

is the reserved subnet broadcast address. The usable host address range is

172.16.10.17

through

172.16.10.30

. Therefore,

172.16.10.17

and

172.16.10.28

are the two valid host addresses.

Step-by-Step Solution

Determine the subnet block size from the prefix length.

A /28 prefix corresponds to a subnet mask of 255.255.255.240. The block size in the 4th octet is

256 - 240 = 16

Knowing the block size helps establish the boundaries of the subnet.

Calculate the network address, broadcast address, and usable host range.

Network address = 172.16.10.16; Broadcast address =

172.16.10.16 + 16 - 1 = 172.16.10.31

; Usable host range = 172.16.10.17 to 172.16.10.30.

The first address in the subnet is reserved for the network ID, and the last address is reserved for the broadcast address.

Evaluate each given option against the usable host range.

172.16.10.17 and 172.16.10.28 fall within 172.16.10.17-172.16.10.30, whereas 172.16.10.16 is the network ID and 172.16.10.31 is the broadcast address.

Only IP addresses strictly between the network ID and broadcast address are assignable to host interfaces.

Key Concept

IPv4 Subnet Boundaries and Usable Host Address Calculation

Question 265Question →

An enterprise organization is expanding its campus wireless infrastructure by deploying lightweight access points (LAPs) across several buildings. To simplify administration, a network engineer installs a central device to handle radio frequency (RF) management, client authentication policies, and control-plane processing for all LAPs. Which network component performs these centralized control functions?

Wireless LAN Controller (WLC)

Layer 2 Switch

Type 2 Hypervisor

Link Layer Discovery Protocol (LLDP) Agent

Show answer & explanation

Answer: Wireless LAN Controller (WLC)

Answer

The Wireless LAN Controller (WLC) is the network component responsible for centralized control-plane management, security policy enforcement, and RF coordination in a lightweight wireless architecture.

In a centralized Cisco wireless network using Split-MAC architecture, the Wireless LAN Controller (WLC) is dedicated to handling heavy control-plane operations, including RF channel and power management, client authentication, roaming assistance, and LAP configuration management.

Step-by-Step Solution

Analyze the operational requirements described in the scenario.

The target component must centralize RF management, client authentication, and control-plane processing for lightweight access points.

Split-MAC architecture divides wireless responsibilities between access points and a central control entity.

Evaluate candidate network components against these requirements.

The Wireless LAN Controller (WLC) specifically terminates CAPWAP control tunnels from LAPs to deliver centralized management, channel assignment, power tuning, and security enforcement.

LAPs rely on the WLC for control-plane logic, whereas Layer 2 switches focus on frame switching, hypervisors manage virtual machines, and discovery protocols handle neighbor advertising.

Key Concept

Wireless LAN Controller (WLC) Role in Split-MAC Architecture

Question 266Question →

An enterprise network administrator is configuring network isolation for virtual machines hosted on a single bare-metal server. The requirement states that traffic from different virtual machine groups must be segregated into distinct 802.1Q VLANs before traversing the server's shared physical uplink. Which component within the virtualization architecture is primarily responsible for performing Layer 2 frame switching between virtual machines and applying the required VLAN tags?

The virtual switch (vSwitch) operating within the hypervisor

The virtual network interface card (vNIC) assigned to the guest operating system

The physical network interface card (pNIC) hardware ASIC controller

The underlying host operating system protocol driver stack

Show answer & explanation

Answer: The virtual switch (vSwitch) operating within the hypervisor

Answer

The virtual switch (vSwitch) operating within the hypervisor is responsible for Layer 2 switching and 802.1Q VLAN tagging for hosted virtual machines.

The virtual switch (vSwitch) is a software program running inside the hypervisor that functions as a Layer 2 switch. It maintains MAC tables, performs local frame forwarding between VMs on the same physical host, and tags/untags 802.1Q Ethernet frames for traffic leaving via physical uplinks.

Step-by-Step Solution

Identify the component that connects virtual machine virtual interfaces (vNICs) to host physical interfaces (pNICs).

The hypervisor implements a software-based Layer 2 virtual switch (vSwitch).

Virtual machines require network connectivity both locally within the host and externally.

Analyze which component handles Layer 2 encapsulation and forwarding features such as VLAN tagging.

The vSwitch manages port groups, maintains a virtual MAC address table, and inserts/strips 802.1Q tags on traffic.

This offloads network control logic from the individual guest operating systems while isolating VM traffic before reaching physical switches.

Key Concept

Virtual Switch (vSwitch) Operation and Functions

Question 267Question →

A network architect is designing a two-tier spine-leaf data center fabric. Each of the 8 leaf switches connects to 48 edge servers using 10 Gbps access ports, generating $480\text{ Gbps}$ of potential access bandwidth per switch. The design policy mandates a maximum bandwidth oversubscription ratio of $3:1$ across the spine fabric. If all spine and leaf interconnect interfaces operate at 40 Gbps, which architecture satisfies the oversubscription constraint while adhering to standard Clos topology rules?

Deploy 4 spine switches and connect each leaf switch with a single 40 Gbps link to every spine switch.

Deploy 2 spine switches and connect each leaf switch with dual 40 Gbps uplinks to every spine switch while trunking the spine switches together.

Deploy 3 spine switches and connect each leaf switch to two designated spine switches to isolate traffic domains.

Deploy 8 spine switches and connect each leaf switch to one dedicated spine switch to eliminate collision domains.

Show answer & explanation

Answer: Deploy 4 spine switches and connect each leaf switch with a single 40 Gbps link to every spine switch.

Answer

Deploying 4 spine switches where each leaf switch has a single 40 Gbps uplink to every spine switch correctly provides 160 Gbps of uplink capacity per leaf (meeting the 3:1 oversubscription ratio) and strictly obeys Spine-Leaf bipartite graph connectivity rules.

Each leaf switch requires

480\text{ Gbps} / 3 = 160\text{ Gbps}

of uplink bandwidth to achieve a

3:1

oversubscription ratio. Using 40 Gbps links, each leaf switch needs 4 uplinks. Standard Spine-Leaf design rules dictate that every leaf switch must connect to every spine switch, and spine switches must not connect to one another. Therefore, deploying 4 spine switches with 1 link from each leaf to each spine satisfies both bandwidth and topological requirements.

Step-by-Step Solution

Calculate total access bandwidth per leaf switch

48 \times 10\text{ Gbps} = 480\text{ Gbps}

access bandwidth per leaf switch.

Determining total ingress/egress access capacity is required to calculate uplink oversubscription.

Calculate required uplink bandwidth based on the 3:1 oversubscription ratio limit

\text{Required Uplink Bandwidth} = \frac{480\text{ Gbps}}{3} = 160\text{ Gbps}

3:1

oversubscription ratio means uplink bandwidth must be at least one-third of the access bandwidth.

Determine the number of 40 Gbps uplinks required per leaf switch

\frac{160\text{ Gbps}}{40\text{ Gbps}} = 4

uplink connections per leaf switch.

Interconnect ports run at 40 Gbps, so 4 ports are required to achieve 160 Gbps total uplink capacity.

Apply Clos (Spine-Leaf) architectural connectivity rules

Deploy 4 spine switches and connect each leaf switch to every spine switch using exactly one 40 Gbps uplink.

In a 2-Tier Spine-Leaf topology, every leaf connects to every spine, spines do not connect to spines, and leaves do not connect to leaves.

Key Concept

Spine-Leaf (Clos) Topology Oversubscription and Fabric Interconnect Rules

Estimated Time:2m 0s

Question 268Question →

Switch SW1 receives a unicast frame on ingress port GigabitEthernet0/2. The frame contains a source MAC address of 7069.5a12.1234 and a destination MAC address of 0011.2233.4455. The MAC address table of SW1 currently has no entries. Which two actions does the switch perform upon receiving this frame? (Select two.)

Select all that apply

The switch records the source MAC address 7069.5a12.1234 and ingress port GigabitEthernet0/2 in its MAC address table.

The switch forwards the frame directly to its configured default gateway to resolve the destination address.

The switch floods the frame out all active ports assigned to the same VLAN, excluding port GigabitEthernet0/2.

The switch floods the frame out all active ports assigned to the VLAN, including port GigabitEthernet0/2.

The switch drops the frame immediately because the destination MAC address is missing from its MAC address table.

Show answer & explanation

Answer: The switch records the source MAC address 7069.5a12.1234 and ingress port GigabitEthernet0/2 in its MAC address table.; The switch floods the frame out all active ports assigned to the same VLAN, excluding port GigabitEthernet0/2.

Answer

The switch records the source MAC address on ingress port GigabitEthernet0/2 in its MAC address table and floods the frame out all operational ports in the same VLAN except GigabitEthernet0/2.

When an Ethernet switch receives a frame, it executes two key processes: learning and forwarding. First, it reads the source MAC address (7069.5a12.1234) and maps it to the receiving interface (GigabitEthernet0/2) in its MAC address table. Second, because the destination MAC address (0011.2233.4455) is not listed in its table, the frame is treated as an unknown unicast frame. The switch floods the frame out every operational port in the VLAN except the ingress port where the frame was received.

Step-by-Step Solution

Examine the ingress frame's source MAC address and ingress interface.

The switch adds an entry for source MAC 7069.5a12.1234 mapped to port GigabitEthernet0/2.

Layer 2 switches perform dynamic MAC learning by recording the source address of every incoming frame.

Examine the ingress frame's destination MAC address against the MAC address table.

Destination MAC 0011.2233.4455 is not found in the table, identifying the frame as an unknown unicast frame.

An empty MAC table contains no matching entry for the destination.

Apply Layer 2 forwarding rules for an unknown unicast frame.

The frame is flooded out all active ports in the same VLAN except ingress port GigabitEthernet0/2.

Flooding ensures the unknown destination host receives the frame without retransmitting traffic out the ingress port.

Key Concept

L2 Switching Concepts and MAC Table Operation

Estimated Time:1m 0s

Question 269Question →

A Cisco Catalyst switch receives an Ethernet frame on access port GigabitEthernet0/1 configured in VLAN 10. The frame contains a source MAC address of 0011.2233.4455 and a unicast destination MAC address of 00aa.bbcc.dde1. The switch currently maintains the following dynamic MAC address table entries:

VLAN	MAC Address	Type	Port
10	0011.2233.4455	DYNAMIC	GigabitEthernet0/2
10	00aa.bbcc.dde1	DYNAMIC	GigabitEthernet0/3
20	00aa.bbcc.dde1	DYNAMIC	GigabitEthernet0/4

Which set of actions will the switch take upon processing this ingress frame?

Update the VLAN 10 entry for 0011.2233.4455 to GigabitEthernet0/1 and forward the frame out GigabitEthernet0/3 only.

Update the VLAN 10 entry for 0011.2233.4455 to GigabitEthernet0/1 and flood the frame out all active ports in VLAN 10 except GigabitEthernet0/1.

Forward the frame out GigabitEthernet0/4 because it has a higher-numbered port matching destination 00aa.bbcc.dde1 without updating the MAC table.

Drop the frame immediately because source MAC address 0011.2233.4455 is already bound to GigabitEthernet0/2 in the MAC table.

Show answer & explanation

Answer: Update the VLAN 10 entry for 0011.2233.4455 to GigabitEthernet0/1 and forward the frame out GigabitEthernet0/3 only.

Answer

Update the VLAN 10 entry for 0011.2233.4455 to GigabitEthernet0/1 and forward the frame out GigabitEthernet0/3 only.

When an Ethernet frame enters a switch port, the switch inspects the source MAC address to maintain its MAC address table. Since MAC 0011.2233.4455 was previously recorded on GigabitEthernet0/2 but arrived on GigabitEthernet0/1 in VLAN 10, the switch updates the table entry to point to GigabitEthernet0/1. Next, the switch inspects the destination MAC address 00aa.bbcc.dde1 within VLAN 10. Finding a matching dynamic entry for GigabitEthernet0/3 in VLAN 10, the switch forwards the frame exclusively out GigabitEthernet0/3.

Step-by-Step Solution

Examine the source MAC address and ingress port/VLAN context.

The source MAC address 0011.2233.4455 arrives on GigabitEthernet0/1 in VLAN 10. The current MAC table lists this address on GigabitEthernet0/2 for VLAN 10.

When a host moves or changes ports, receiving a frame on a new port triggers the switch to update (re-learn) the MAC address table entry for that specific VLAN.

Perform a destination MAC address lookup within the ingress VLAN.

The destination MAC address 00aa.bbcc.dde1 is present in the MAC table under VLAN 10 associated with port GigabitEthernet0/3.

Switches isolate forwarding decisions per VLAN. The matching entry in VLAN 20 on GigabitEthernet0/4 is ignored.

Determine the forwarding action.

The switch performs known unicast forwarding by transmitting the frame out GigabitEthernet0/3 only.

Since the destination MAC address is known within VLAN 10, no unknown unicast flooding is required.

Key Concept

Layer 2 MAC Table Learning, Station Mobility, and Per-VLAN Forwarding Decisions

Estimated Time:2m 0s

Question 270Question →

A network administrator configures a router interface using EUI-64 address generation with the command `ipv6 address 2001:db8:100:1::/64 eui-64`. If the physical MAC address of the interface is `bc24.1122.3344`, which 64-bit interface identifier will be automatically derived?

BE24:11FF:FE22:3344

BC24:11FF:FE22:3344

BE24:11FF:FF22:3344

BE24:1122:33FF:FE44

Show answer & explanation

Answer: BE24:11FF:FE22:3344

Answer

The derived interface identifier is BE24:11FF:FE22:3344.

The 64-bit EUI-64 interface identifier is generated by taking the 48-bit MAC address (`bc24.1122.3344`), inserting the 16-bit value `FFFE` in the middle (between `11` and `22`), and flipping the 7th bit of the first byte. The binary representation of `BC` is `10111100`; flipping the 7th bit yields `10111110`, which equals `BE` in hexadecimal format. Combining these results yields `BE24:11FF:FE22:3344`.

Step-by-Step Solution

Split the 48-bit MAC address into two 24-bit halves.

Left half: `BC-24-11`, Right half: `22-33-44`.

EUI-64 standard requires inserting a 16-bit value between the Organizationally Unique Identifier (OUI) and the NIC-specific portion.

Insert the 16-bit hexadecimal value FFFE into the midpoint.

`BC-24-11-FF-FE-22-33-44`.

Expands the 48-bit MAC address to a 64-bit structure.

Invert the 7th bit (Universal/Local bit) of the first byte.

First byte `BC` (binary `10111100`) becomes `BE` (binary `10111110`). Combining all parts in IPv6 format yields `BE24:11FF:FE22:3344`.

In EUI-64 addresses, setting the 7th bit to 1 specifies that the interface ID is globally unique.

Key Concept

EUI-64 Interface Identifier Generation

Question 271Question →

A network administrator needs to perform the initial out-of-band configuration of a newly unboxed Cisco Catalyst switch by connecting a laptop directly to the switch RJ-45 console port. Which cable type must be used to establish this management connection?

Rollover cable

Straight-through cable

Crossover cable

Single-mode fiber patch cable

Show answer & explanation

Answer: Rollover cable

Answer

Rollover cable

A rollover cable is designed specifically for out-of-band management access to Cisco networking device console ports. It completely reverses the pin order from pin 1 through 8 on one connector to pins 8 through 1 on the opposing RJ-45 connector, allowing serial communication between a terminal/laptop and the device management hardware.

Step-by-Step Solution

Identify the purpose of the interface link

The objective is local out-of-band administrative access through the switch's RJ-45 console interface.

Console management uses serial communication signal pinouts rather than standard Ethernet network data pinouts.

Determine the required cabling pinout structure

A rollover cable (also known as a Yost cable or console cable) reverses the pin ordering from end to end (pin 1 to pin 8, pin 2 to pin 7, etc.), aligning the terminal adapter serial signals with the Cisco console port.

Standard Ethernet cable pinouts (T568A or T568B straight-through and crossover cables) do not map the required serial transmit and receive signals correctly.

Key Concept

Cisco Console Port Cabling and Rollover Pinouts

Question 272Question →

An application architect is designing a lightweight heartbeat monitoring service for IoT sensors. To conserve network bandwidth over low-power links, the design requires minimizing transport layer protocol header overhead for small status payloads. Which transport layer characteristic makes UDP the most suitable choice for this application?

UDP guarantees in-order delivery of telemetry packets using sequence numbers without requiring acknowledgment packets.

UDP utilizes a fixed 8-byte header and operates without connection setup overhead.

UDP provides flow control through a dynamic sliding window mechanism to adjust transmission rates on constrained links.

UDP reduces transmission overhead by dynamically negotiating a compressed 20-byte transport header during session establishment.

Show answer & explanation

Answer: UDP utilizes a fixed 8-byte header and operates without connection setup overhead.

Answer

UDP utilizes a fixed 8-byte header and operates without connection setup overhead.

UDP features a minimal, fixed 8-byte header (containing source port, destination port, length, and checksum) and operates connectionlessly without handshake or state management overhead, making it ideal for bandwidth-constrained, small-payload messaging.

Step-by-Step Solution

Analyze the application requirements from the scenario.

The application requires minimal transport layer header overhead for small status messages over low-bandwidth links.

Choosing between TCP and UDP depends on balancing reliability requirements against protocol overhead.

Compare TCP and UDP transport header sizes and connection overhead.

TCP uses a minimum 20-byte header and requires a 3-way handshake, whereas UDP uses a fixed 8-byte header and is connectionless.

UDP's 8-byte header minimizes payload ratio overhead significantly compared to TCP's 20-byte minimum header.

Select the option that correctly describes UDP's low-overhead characteristics.

The statement highlighting UDP's fixed 8-byte header and lack of connection setup overhead is correct.

This directly satisfies the requirement for low-power link efficiency.

Key Concept

UDP vs TCP Header Overhead and Connection Mechanics

Question 273Question →

A network administrator is designing an IPv4 addressing scheme for a local subnet that must support $60$ usable host IP addresses. The administrator assigns the smallest subnet prefix length (largest mask) that accommodates this host requirement. How many total usable host IPv4 addresses are provided by this assigned subnet?

Show answer & explanation

Answer: 62

Answer

The assigned subnet provides 62 total usable host IPv4 addresses.

To support 60 host addresses, a subnet must allocate 6 host bits because

2^6 - 2 = 62

usable addresses. 5 host bits only provide 30 usable addresses (

2^5 - 2

), which is insufficient.

Step-by-Step Solution

Determine the required number of host bits.

6 host bits are required (

2^6 - 2 = 62

5 host bits only yield

2^5 - 2 = 30

usable IP addresses, which is not enough for 60 hosts.

Calculate the usable host capacity of the subnet.

62 usable host addresses.

With 6 host bits, the total IP space is

2^6 = 64

. Subtracting the network address and the broadcast address leaves 62 usable host IP addresses.

Key Concept

IPv4 Usable Host Calculation

Estimated Time:45s

Question 274Question →

A network engineer is deploying IPv6 on a Cisco router interface with the burned-in MAC address `b496.910a.bcde`. The interface is enabled to automatically formulate its link-local address using standard EUI-64 formatting. What complete, compressed link-local IPv6 address will be generated on this interface?

Show answer & explanation

Answer: fe80::b696:91ff:fe0a:bcde; FE80::B696:91FF:FE0A:BCDE; fe80:0000:0000:0000:b696:91ff:fe0a:bcde; FE80:0000:0000:0000:B696:91FF:FE0A:BCDE

Answer

fe80::b696:91ff:fe0a:bcde

The MAC address `b496.910a.bcde` consists of octets `b4`, `96`, `91`, `0a`, `bc`, `de`. Inverting the 7th bit of `b4` (binary `1011 0100`) gives binary `1011 0110` (hex `b6`). Inserting `fffe` into the midpoint between `91` and `0a` creates the 64-bit interface ID `b696:91ff:fe0a:bcde`. Combining this with the IPv6 link-local prefix `fe80::/64` yields `fe80::b696:91ff:fe0a:bcde`.

Step-by-Step Solution

Split the 48-bit MAC address into two 24-bit halves.

Left half: `b4-96-91`, Right half: `0a-bc-de`.

The 64-bit interface identifier is constructed by inserting a 16-bit hex delimiter into the middle of the 48-bit MAC address.

Invert the Universal/Local (U/L) bit (the 7th bit of the first byte).

The first byte `b4` (hex `1011 0100` in binary) becomes `1011 0110` (hex `b6`).

In EUI-64 conversion, the 7th bit (U/L bit) of the first byte must be inverted (0 becomes 1 for globally unique MAC addresses).

Insert the 16-bit hexadecimal string `FFFE` between the two halves.

The 64-bit interface identifier becomes `b696:91ff:fe0a:bcde`.

Inserting `FFFE` expands the 48-bit MAC address into a valid 64-bit EUI-64 interface identifier.

Prepend the IPv6 link-local prefix (`fe80::/64`).

The full link-local address is `fe80::b696:91ff:fe0a:bcde`.

IPv6 link-local addresses use the reserved prefix `fe80::/10` with standard interface configuration defaulting to `fe80::/64` followed by the 64-bit EUI-64 interface ID.

Key Concept

IPv6 EUI-64 Link-Local Address Derivation

Estimated Time:2m 0s

Question 275Question →

A network architect is designing a virtualized data center infrastructure utilizing bare-metal hypervisors and virtual switches. Which two statements accurately describe the roles and functions of these components? (Choose two.)

Select all that apply

The Type 1 hypervisor runs directly on the underlying server hardware without requiring a host operating system.

The virtual switch handles Layer 2 frame forwarding between virtual machines configured on the same host.

A Type 1 hypervisor operates as an application layer service installed on top of an existing general-purpose operating system.

A virtual switch inherently divides broadcast domains at Layer 3 without needing VLAN configurations or external routers.

Show answer & explanation

Answer: The Type 1 hypervisor runs directly on the underlying server hardware without requiring a host operating system.; The virtual switch handles Layer 2 frame forwarding between virtual machines configured on the same host.

Answer

The correct statements are that the Type 1 hypervisor runs directly on physical server hardware without requiring a host operating system, and that the virtual switch handles Layer 2 frame forwarding between virtual machines on the same host.

Type 1 hypervisors install directly onto physical hardware without requiring a host OS, providing efficient allocation of hardware resources to virtual machines. The internal virtual switch operates at Layer 2 to manage frame forwarding between virtual machines connected to the same virtual network.

Step-by-Step Solution

Analyze the architecture of Type 1 hypervisors

Confirm that Type 1 hypervisors (bare-metal) install directly on the hardware platform without an intermediate operating system.

Direct access to physical hardware provides higher efficiency and lower processing overhead for enterprise virtualization.

Examine the role of software virtual switches inside hypervisors

Confirm that virtual switches perform Layer 2 forwarding for guest virtual machine network traffic on the host.

Virtual switches mimic physical Layer 2 switches to connect virtual network interface cards (vNICs).

Evaluate distractor statements for conceptual misconceptions

Identify that operating on top of a host OS is characteristic of Type 2 hypervisors, and that virtual switches operate at Layer 2 rather than separating broadcast domains at Layer 3 automatically.

Hypervisor classification and OSI layer responsibilities must be strictly distinguished.

Key Concept

Hypervisor Types and Virtual Switch Operations

Question 276Question →

A network architect is designing a Variable Length Subnet Mask (VLSM) addressing scheme for a enterprise branch using the allocated summary block $172.16.48.0/20$ . Subnets must be allocated sequentially starting from the beginning of the block, prioritized from the largest host requirement to the smallest, to preserve contiguous unassigned space. The host requirements for the subnets are as follows:

- Department Alpha: $120$ hosts
- Department Beta: $60$ hosts
- Department Gamma: $28$ hosts
- Department Delta: $12$ hosts

Which IPv4 network address and subnet mask must be assigned to Department Gamma?

172.16.48.192

with subnet mask

255.255.255.224

172.16.48.128

with subnet mask

255.255.255.224

172.16.48.192

with subnet mask

255.255.255.240

172.32.48.192

with subnet mask

255.255.255.224

Show answer & explanation

Answer: $172.16.48.192$ with subnet mask $255.255.255.224$

Answer

Department Gamma should be assigned network address 172.16.48.192 with subnet mask 255.255.255.224.

The correct response identifies 172.16.48.192 255.255.255.224 as the network address and subnet mask. Calculating sequentially from the summary block 172.16.48.0/20: Department Alpha uses 172.16.48.0/25 (128 addresses, 126 usable), ending at 172.16.48.127. Department Beta uses 172.16.48.128/26 (64 addresses, 62 usable), ending at 172.16.48.191. Department Gamma requires 28 usable hosts, which fits in a /27 subnet (32 addresses, 30 usable, mask 255.255.255.224) starting at the next boundary: 172.16.48.192.

Step-by-Step Solution

Determine prefix length and block size for Department Alpha (120 hosts)

Requires 120 hosts. Formula

2^n - 2 \ge 120 \implies n=7

host bits. Subnet mask /25 (255.255.255.128, block size 128). Range: 172.16.48.0 to 172.16.48.127.

Always allocate largest subnets first to optimize contiguous IP space.

Determine prefix length and block size for Department Beta (60 hosts)

Requires 60 hosts. Formula

2^n - 2 \ge 60 \implies n=6

host bits. Subnet mask /26 (255.255.255.192, block size 64). Next available boundary is 172.16.48.128. Range: 172.16.48.128 to 172.16.48.191.

Department Beta starts immediately after Department Alpha's broadcast address.

Calculate network address and subnet mask for Department Gamma (28 hosts)

Requires 28 hosts. Formula

2^n - 2 \ge 28 \implies n=5

host bits (

2^5 - 2 = 30

usable hosts). Subnet mask /27 (255.255.255.224, block size 32). Next available boundary is 172.16.48.192.

Department Gamma starts at 172.16.48.192/27, providing 30 usable host IPs (172.16.48.193 - 172.16.48.222).

Key Concept

Variable Length Subnet Masking (VLSM) and Host Capacity Calculation

Estimated Time:2m 0s

Question 277Question →

A network engineer is evaluating cabling standards and optical transceiver specifications for a campus network refresh. When connecting switch interfaces across various media types, which two statements correctly describe physical interface and cabling characteristics according to Ethernet standards? (Select two.)

Select all that apply

10GBASE-SR transceivers operate over multimode fiber using an 850 nm wavelength and can reach up to 300 meters on OM3 cabling.

1000BASE-T Ethernet requires all four wire pairs in Category 5e or higher UTP cabling to operate concurrently in full-duplex mode.

100BASE-TX crossover cables swap pins 1 and 2 with pins 4 and 5 at the opposite RJ-45 connector while keeping pins 3 and 6 straight through.

Single-mode fiber (SMF) cores typically measure 50 to 62.5 microns in diameter, resulting in high modal dispersion compared to multimode fiber (MMF).

1000BASE-LX transceivers use 1310 nm long-wavelength lasers and are strictly restricted to single-mode fiber infrastructure.

Show answer & explanation

Answer: 10GBASE-SR transceivers operate over multimode fiber using an 850 nm wavelength and can reach up to 300 meters on OM3 cabling.; 1000BASE-T Ethernet requires all four wire pairs in Category 5e or higher UTP cabling to operate concurrently in full-duplex mode.

Answer

The correct statements are that 10GBASE-SR transceivers operate over multimode fiber using an 850 nm wavelength reaching up to 300 meters on OM3 cabling, and 1000BASE-T Ethernet requires all four wire pairs in Category 5e or higher UTP cabling to operate concurrently.

10GBASE-SR transceivers specify short-wavelength 850 nm light over multimode fiber with standard distance caps of 300 meters on OM3 and 400 meters on OM4. Additionally, 1000BASE-T requires all four copper pairs in Cat5e or better cables, transmitting full-duplex signals over all eight conductors simultaneously.

Step-by-Step Solution

Analyze optical fiber transceiver specifications

10GBASE-SR (Short Reach) uses 850 nm lasers over multimode fiber (MMF). It supports up to 300 m on laser-optimized OM3 MMF and 400 m on OM4 MMF.

Matching transceiver naming conventions (SR vs LR vs ER) to light wavelengths and distance limits is critical for optical link design.

Analyze twisted-pair copper transmission mechanics

1000BASE-T achieves 1 Gbps throughput across Category 5e/6 UTP by utilizing hybrid circuits that allow bi-directional transmission on all four pairs simultaneously.

10BASE-T and 100BASE-TX only use 2 pairs (4 wires), whereas Gigabit 1000BASE-T requires all 4 pairs (8 wires).

Evaluate and eliminate incorrect distractor claims regarding pinouts, fiber core diameters, and optic compatibility

Crossover cables swap pins 1/2 with pins 3/6. Single-mode fiber has a smaller core (9 microns) with no modal dispersion. 1000BASE-LX can run on MMF with a mode-conditioning patch cable.

Distinguishing between MMF and SMF physical attributes and recognizing optic deployment exceptions prevents physical layer design errors.

Key Concept

Physical Interface Standards and Cabling Specifications

Estimated Time:2m 0s

Question 278Question →

When deriving an IPv6 64-bit interface identifier using the EUI-64 process from a 48-bit MAC address, which bit position (counted from left to right within the first octet) is inverted?

Show answer & explanation

Answer: 7; 7th; 7th bit; bit 7; seventh; seventh bit

Answer

The 7th bit position (the Universal/Local bit) in the first octet of the MAC address is inverted.

During the EUI-64 calculation process, the 7th bit from the left of the MAC address's first octet (the Universal/Local bit) is inverted. For example, a universal MAC address starting with binary 00000000 (0x00) becomes 00000010 (0x02) after the 7th bit is flipped from 0 to 1.

Step-by-Step Solution

Identify the modification rules for converting a 48-bit MAC address to an EUI-64 64-bit interface identifier.

The two required modifications are inserting the 16-bit value

FFFE

into the middle of the MAC address and inverting the Universal/Local (U/L) bit.

EUI-64 formatting requires extending the 48-bit MAC address to 64 bits and adjusting the scope indicator.

Determine the exact location of the Universal/Local (U/L) bit.

The U/L bit is located at the 7th bit position from the left in the first octet of the MAC address.

In standard IEEE MAC address formatting, bit 7 specifies whether the address is universally administered (0) or locally administered (1).

Key Concept

EUI-64 Universal/Local Bit Inversion

Question 279Question →

A network administrator is evaluating transport layer protocol characteristics for two distinct enterprise applications: a database replication service requiring guaranteed in-order data transfer, and a emergency voice alert system prioritizing minimum transmission delay. Which statements correctly distinguish how TCP and UDP support these requirements? (Select TWO.)

Select all that apply

TCP utilizes sequence numbers and acknowledgments to ensure reliable, ordered byte-stream delivery for data replication.

UDP provides flow control for voice alerts by dynamically adjusting sender window size based on buffer availability.

UDP operates with a fixed 8-byte header and avoids connection handshake latency, making it suitable for real-time voice delivery.

TCP reduces protocol overhead by operating in a connectionless mode during large database file transfers.

Show answer & explanation

Answer: TCP utilizes sequence numbers and acknowledgments to ensure reliable, ordered byte-stream delivery for data replication.; UDP operates with a fixed 8-byte header and avoids connection handshake latency, making it suitable for real-time voice delivery.

Answer

TCP guarantees reliable in-order delivery using sequence numbers and acknowledgments, while UDP provides low overhead with an 8-byte header and connectionless transmission ideal for latency-sensitive applications.

TCP is connection-oriented and uses sequence numbers along with acknowledgment numbers to track stream state and guarantee in-order delivery. Conversely, UDP is connectionless and lightweight, utilizing a small 8-byte header to eliminate transmission overhead for latency-sensitive traffic such as voice broadcasts.

Step-by-Step Solution

Analyze transport protocol requirements for the database replication service.

Database replication requires absolute data accuracy, loss detection, and in-order delivery, which demands TCP features like sequencing, ACKs, and retransmission.

Data integrity takes priority over transport latency.

Analyze transport protocol requirements for the voice alert system.

Voice traffic is sensitive to delay and jitter but can tolerate minor packet loss, making UDP the ideal choice due to its low 8-byte overhead and lack of handshake delays.

Connectionless best-effort delivery prevents head-of-line blocking.

Evaluate protocol features against provided options.

The options describing TCP sequencing/ACK reliability mechanics and UDP minimal 8-byte header/connectionless operation accurately state protocol behaviors.

Flow control and connectionless bulk transfer assertions misplace TCP and UDP mechanics.

Key Concept

TCP vs UDP Characteristics and Header Comparison

Question 280Question →

An administrator is configuring a switch management VLAN interface with an IPv4 address of 172.16.5.33 and a subnet mask of 255.255.255.224 (/27). What is the total number of usable host IP addresses available on this subnet?

Show answer & explanation

Answer: 30

Answer

The total number of usable host IP addresses available on a /27 subnet is 30.

The correct answer is 30. For a prefix of /27 (subnet mask 255.255.255.224), there are

32 - 27 = 5

host bits available. The total address block size is

2^5 = 32

. Subtracting the network identifier and the broadcast address (

32 - 2

) gives exactly 30 usable host IP addresses.

Step-by-Step Solution

Determine the number of host bits from the prefix length.

32 total IPv4 address bits - 27 network bits = 5 host bits (

n = 5

The prefix length /27 defines 27 bits for the network portion, leaving the remaining bits for host addressing.

Calculate the total number of IP addresses in the subnet block.

2^5 = 32

total IP addresses.

The formula

2^n

calculates the total number of binary combinations available with

n

host bits.

Subtract the non-assignable network and broadcast addresses to find the usable host count.

32 - 2 = 30

usable host IP addresses.

In standard IPv4 subnetting, the first address (network ID) and last address (directed broadcast address) cannot be assigned to individual hosts.

Key Concept

Calculating Usable IPv4 Host Addresses using the

2^n - 2

Formula

Estimated Time:45s

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